Need help with Lagrange Multipliers I need to maximize $U = BM$ with constraits:
$6B +3M = 60$, $B>0$ and $M>0$.
The Lagrange function is $L=U + \lambda (6B+3M-60) + KB + HM$.
So
$$\partial_{\lambda}L= 6B+3M-60=0$$
$$\partial_{K}L = B=0$$
$$\partial_{H}L = M=0$$
$$\partial_{M}L = \partial_{M} (BM) +\lambda(3+6\partial_{M}B)+K\partial_{M}B +H=((-0.5)M+B)-0.5K+H$$
where $B=\frac{60-3M}{6}$ so $\partial_{M}B = -\frac{1}{2}$. But I am confused here by constrait $B=0$ and $M=0$, they cannot be. How can I make sure $B$ and $M$ are non-negative?
 A: Edit: I have finally figured out what you were trying to do. The method of Lagrange multipliers incorporates constraints that are equalities, not inequalities. What you tried to do was, in effect, tack on the extra conditions $B=0$ and $M=0$ to the problem, which makes it inconsistent. When you deal with inequalities you simply throw out whatever solutions to $\nabla\Lambda=0$ do not satisfy them.
A solution should go like the following (I have changed the numbers slightly so that it is not the same exact problem).


Problem. Maximize the function $U(B,M)=B\cdot M$ under the constraints $$g(B,M)=B+2M=20,\qquad B,M>0.$$


*

*Step 1. The Lagrange function is $$\Lambda(B,M,\lambda)=U(B,M)+\lambda \cdot (g(B,M)-20)$$ $$=BM+\lambda(B+2M-20).$$


*Step 2. Set $\nabla_{}\Lambda=0$: $$\nabla_{B,M,\lambda}\Lambda(B,M,\lambda)=(0,0,0),\text{ i.e.}$$ $$\frac{\partial\Lambda}{\partial B}=M+\lambda=0;$$ $$\frac{\partial\Lambda}{\partial M}=B+2\lambda=0;$$ $$\frac{\partial\Lambda}{\partial\lambda}=B+2M-20=0.$$


*Step 3. Set $B=-2\lambda,M=-\lambda$, derived from the second and first equations above respectively, then substitute them into the third equation and solve for $\lambda$: $$(-2\lambda)+2(-\lambda)-20=0$$ $$\implies \lambda=-5.$$


*Step 4: Plug in $\lambda=-5$ into $B$ and $M$'s expressions from last step to find $B=10$ and $M=5$.


*Step 5: Check the work. The answer gives two positive values that satisfy the original linear constraint. It gives a maximum value of $U(10,5)=50$, whereas $U$ at the endpoints $B=0$ and $M=0$ simply evaluate to $0$, much lower than $50$, therefore the answer is correct.
